# hwsol4 - Homework 5 Solutions Course Reader Exercises 15.3...

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Homework 5 Solutions Course Reader Exercises 15.3 The matrix for the rotation through π/ 4 radians is A = " 2 2 - 2 2 2 2 2 2 # Since the reflection sends e 1 to - e 2 and e 2 to - e 1 , its matrix is B = 0 - 1 - 1 0 Since T is the composition of these transformations, its matrix is BA = " - 2 2 - 2 2 - 2 2 2 2 # 15.4 First we need to consider the effect T has on the standard basis vectors in R 3 . Since the x -axis is fixed by T , T ( e 1 ) = e 1 . The figure below illustrates the effect T has on the y - z plane. y z e e e e T( ) T( ) 2 2 3 3 The x components of both T ( e 2 ) and T ( e 3 ) are zero. Using a little trigonometry, the y and z components of T ( e 2 ) are seen to be 3 / 2 and 1 / 2, respectively. Likewise the y and z components of T ( e 3 ) are - 1 / 2 and 3 / 2, respectively. Thus T ( e 1 ) = 1 0 0 T ( e 2 ) = 0 3 / 2 1 / 2 T ( e 3 ) = 0 - 1 / 2 3 / 2 so the matrix for T is A = 1 0 0 0 3 / 2 - 1 / 2 0 1 / 2 3 / 2 1

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Since S leaves the z -axis fixed, S ( e 3 ) = e 3 . The effect of S on the x - y plane is illustrated below. x y e e e e T( ) T( ) 1 1 2 2 The z components of both S ( e 1 ) and S ( e 2 ) are zero. The x and y components of S ( e 1 ) are both 2 / 2 and the x and y components of S ( e 2 ) are - 2 / 2 and 2 / 2, respectively. Thus S ( e 1 ) = 2 / 2 2 / 2 0 S ( e 2 ) = - 2 / 2 2 / 2 0 S ( e 3 ) = 0 0 1 so the matrix for S is B = 2 / 2 - 2 / 2 0 2 / 2 2 / 2 0 0 0 1 The matrix for T S is AB = 2 / 2 - 2 / 2 0 6 / 4 6 / 4 - 1 / 2 2 / 4 2 / 4 3 / 2 The matrix for S T is BA = 2 / 2 - 6 / 4 2 / 4 2 / 2 6 / 4 - 2 / 4 0 1 / 2 3 / 2 16.11 Remember that I n A = AI n = A . Now expand ( I n + A ) ( I n - A ) = I n I n - I n A + AI n - AA = I n - AA = I n - A 2 16.12 Be careful. AB 6 = BA ! So we expand ( B + A ) ( B - A ) = BB - BA + AB - AA 2
16.13 Saying that A = A - 1 is the same as saying that AA = I : a b b - a a b b - a = 1 0 0 1 a 2 + b 2 0 0 a 2 + b 2 = 1 0 0 1 So we deduce that a 2 + b 2 = 1. 16.14 For the matrix to have an inverse, its reduced row echelon from must be the identity – there should be no free variables. So let’s try and row reduce as best as we can. R 2 - R 1 1 1 1 0 1 k - 1 1 4 k 2 R 3 - R 1 1 1 1 0 1 k - 1 0 3 k 2 R 1 - R 2 1 0 2 - k 0 1 k - 1 0 3 k 2 R 3 - 3 R 2 1 0 2 - k 0 1 k - 1 0 0 k 2 - 3 k + 2 Now we can ensure that the first two columns are pivot columns. The third column will be a pivot column provided that k 2 - 3 k + 2 6 = 0 – in which case we can use it to eliminate the other entries in the third column. More geometrically, provided that k 2 - 3 k + 2 6 = 0, the vectors corresponding to our three columns will be linearly independent. Note that k 2 - 3 k + 2 is the determinant of our original matrix.

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• Fall '07
• Staff
• Linear Algebra, 0 1 k, 0 0 0 0 Second, 1 0 2-k, 1 1 0 2-k

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